How can I show that $f$ can at most have $3$ $x$-intercepts in the interval $K$?

Question

Let $f$ be the function differenitable $2$ times on the interval $K$. Assume that $f''$ has exactly $1$ $x$-intercept. How can I show that $f$ can at most have $3$ $x$-intercepts in the interval $K$?

Answer 1

Suppose by contradiction that $\{x_1,x_2,x_3,x_4\}\subset K$ with $x_i<x_{i+1}$ and $0=f(x_i)$ for $i\in \{1,2,3,4\}.$

By Rolle's Theorem ( a special case of the Mean Value Theorem) there exist $y_1,y_2,y_3$ with $x_j<y_j<x_{j+1}$ and $0=f'(y_j) $ for $j\in \{1,2,3\}.$

By Rolle's Theorem again, there exist $z_1,z_2$ with $y_k<z_k<y_{k+1}$ and $ 0=f''(z_k)$ for $k\in \{1,2\}.$

But $z_1\ne z_2$ (because $z_1<y_2<z_2$) and $\{z_1,z_2\}\subset [y_1,y_3]\subset [x_1,x_4]\subset K,$ so $f''$ vanishes on at least $2$ members of $K, $ contrary to the hypothesis.

Answer 2

If $h''$ has 1 $x$-intercept, what can you say about $h'$ in that point? how many $x$-intercepts has $h'$ at most?